Question

The volume of the $y=$tethradron formed by the coterminous edges $\overline{a},\overline{b},\overline{c}$ is $3$.Then the volume of the parallelepiped formed by the coterminous edges $\overline{a}+\overline{b},\overline{b}+\overline{c},\overline{c}+\overline{a}$ is

• A. $6$
• B. $18$
• C. $36$
• D. $9$

Answer 1

The correct option is C $36$

Given:

$[\overrightarrow{a}\cdot (\overrightarrow{b}\times \overrightarrow{c})]=3\times 6=18$

Volume $=[(\overrightarrow{a}+\overrightarrow{b}).((\overrightarrow{b}+\overrightarrow{c})\times (\overrightarrow{c}+\overrightarrow{a})\left)\right]$

$=\overrightarrow{a}.(\overrightarrow{b}\times \overrightarrow{c})+\overrightarrow{a}.(\overrightarrow{b}\times \overrightarrow{a})+\overrightarrow{a}.(\overrightarrow{c}\times \overrightarrow{c})+\overrightarrow{a}.(\overrightarrow{c}\times \overrightarrow{a})+\overrightarrow{b}.(\overrightarrow{b}\times \overrightarrow{c})+\overrightarrow{b}.(\overrightarrow{b}\times \overrightarrow{a})+\overrightarrow{b}.(\overrightarrow{c}\times \overrightarrow{c})+\overrightarrow{b}.(\overrightarrow{c}\times \overrightarrow{a})$

$\therefore$ Volume $=\overrightarrow{a}.(\overrightarrow{b}\times \overrightarrow{c})+\overrightarrow{b}.(\overrightarrow{c}\times \overrightarrow{a})=2[\overrightarrow{a}\cdot (\overrightarrow{b}\times \overrightarrow{c})]=2\times 18=36$

Since, $\overrightarrow{a}.(\overrightarrow{b}\times \overrightarrow{c})=\overrightarrow{b}.(\overrightarrow{c}\times \overrightarrow{a})$